Question #47c93

1 Answer
Jan 13, 2016

Answer:

Here's what I got.

Explanation:

Your tool of choice for this problem will be the Arrhenius equation, which establishes a relationship between the rate constant of a reaction, #k#, and the temperature at which it occurs, #T#

#color(blue)(k = A * "exp"(-E_a/(RT))" " " "color(purple)((1))" "#, where

#A# - a pre-exponential factor
#E_a# - the activation energy of the reaction
#R# - the universal gas constant, in this context given as #8.3145"J"/("mol K")#

Now, the Arrhenius parameters are #A# and #E_a#.

Since you're basically dealing with one equation and two unknowns, you will have to use the information given to find a way to eliminate one of the two Arrhenius parameters first.

You can do that by writing the Arrhenius equation in non-exponential form

#ln(k) = ln[A * "exp"(-E_a/(RT))]#

This is equivalent to

#ln(k) = ln(A) + ln["exp"(-E_a/(RT))]#

#ln(k) = ln(A) - E_a/(RT)#

Now, notice that the problem provides you with information about the value of the rate constant at two different temperatures

#T_1 = 273.15 + 24 = "297.15 K"#

#T_2 = 273.15 + 37 = "310.15 K"#

This means that you can write

#ln(k_1) = ln(A) - E_a/(R * T_1) -># for #k_1# and #T_1#

#ln(k_2) = ln(A) - E_a/(R * T_2) -># for #k_2# and #T_2#

Notice that you can get rid of the #ln(A)# term by subtracting the second equation for the first one.

#ln(k_1) - ln(k_2) = color(red)(cancel(color(black)(ln(A)))) - color(red)(cancel(color(black)(ln(A)))) - E_a/R * (1/T_1 - 1/T_2)#

This will get you

#color(blue)(ln(k_1/k_2) = - E_a/R * (1/T_1 - 1/T_2))#

You can now find the activation energy of the reaction - keep in mind that activation energy must carry a positive sign, since it expresses energy required

#E_a = - (R * ln(k_1/k_2))/(1/T_1 - 1/T_2)#

#E_a = - (8.3145 "J"/("mol" * color(red)(cancel(color(black)("K")))) * ln( (1.70 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))/(2.01 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))))/((1/297.15 - 1/310.15) 1/color(red)(cancel(color(black)("K"))))#

#E_a = "9873.5 J/mol"#

To get the value of the pre-exponential factor, pick a value for #k# and #T# and plug in the activation energy into equation #color(purple)((1))#.

For #k_1# and #T_1#, you will get

#k_1 = A * "exp"(-E_a/(RT_1))#

This will get you

#A = k_1/("exp"(-E_a/(RT)))#

Notice that since the exponential has to be unitless, #A# will have the same units as #k#. Plug in your values to get

#A = (1.70 * 10^(-2)"L mol"^(-1)"s"^(-1))/("exp"(- (9873.5 color(red)(cancel(color(black)("J")))/color(red)(cancel(color(black)("mol"))))/(8.3145color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 297.15color(red)(cancel(color(black)("K"))))))#

#A = "0.92475 L mol"^(-1)"s"^(-1)#

I'll leave the answers rounded to three sig figs.

#E_a = color(green)("9.87 kJ/mol") -># expressed in kilojoules per mole

#A = color(green)("0.925 L mol"^(-1)"s"^(-1))#